3

Let $X$ be a set. Recall that an atlas of Banach Manifold on $X$ is a set of charts $\mathcal{A} = \{(U_\alpha, \varphi_\alpha, E_\alpha)\}_\alpha$ where $\varphi_\alpha: U_\alpha\to \varphi_\alpha(U_\alpha)\subset E_\alpha$ is a bijection, $E_\alpha$ is a Banach space, and a compatibility condition between the charts is satisified. This standard definition can be found in the Wikipedia article on Banach Manifolds.

One could so define a Banach Manifold to be a couple $(X, \mathcal{A})$. But a larger atlas is needed to do differential geometry. Namely, all charts that are compatible with $\mathcal{A}$ are also charts. Said differently, if $\mathcal{B}$ is an atlas on $X$ such that $\mathcal{A}\bigcup\mathcal{B}$ is also an atlas on $X$, the two atlas are said to be compatible. So charts of $\mathcal{B}$ are also wanted charts, more precisely $(X, \mathcal{A})$ and $(X, \mathcal{B})$ describe the same manifold.

The collection of all atlases on $X$ is not a set because one can choose a Banach space from the class of all Banach spaces... On Wikipedia, I quote:

Compatibility defines an equivalence relation on the class of all possible atlases on $X$.

Is it possible to define an equivalence relation like that on a proper class? If so, it's not with ZFC, so is it with NBG? Could you give me more details?

Then, still on Wikipedia,

A $C^r$-manifold structure on $X$ is then defined to be a choice of equivalence class of atlases on $X$ of class $C^r$

So let's call $\bar{\mathcal{A}}$ this equivalence class. Now could a Banach manifold be the couple $(X, \bar{\mathcal{A}})$?. I doubt that it's logically consistent to talk about the couple $(X, \bar{\mathcal{A}})$ since $\bar{\mathcal{A}}$ is a proper class. Is it?

I should say that it's not really a question about differential geometry but about logic. Many (most?) of the authors I read about this don't talk about this logical issue. For example, S. Lang's Foundation of Differential Geometry or Bourbaki's "Fascicule des résultats". But Bourbaki seem's to use their own logic.

Finally, I should add that in the case you restrict the possibilities to a set of Banach spaces (for instance $\{\mathbb{R}^n,\, n\geq0\}$), you don't have any issues. But it's more satisfactory to allow every Banach space or classes like Hilbert spaces or finite dimensional normed spaces.

Thanks, Paul.

paeolo
  • 330
  • 1
  • 11
  • 1
    All Banach spaces you could possibly use in an atlas for (connected) $X$ are isomorphic, so there's no trouble when you're trying to specify a single manifold. Specify $X$ by picking a Banach space $E$ and then a collection of charts etc based on open subsets of $E$. –  Jul 04 '17 at 23:18
  • It's true, but suppose $X_E$ is your manifold with $E$-valued charts and underlying set $X$. Let $\phi: E\to F$ be an isomorphism of Banach spaces. Then composing all charts with $\phi$ will lead to another atlas with values in $F$. Then endowing $X$ with this $F$-valued atlas will give a different manifold $X_F$. They are really the same (from the point of view of differential geometry), but they are logically (i mean set-theoretically) different. I would prefer a logic solution that makes $X_E$ and $X_F$ the same (set-theoretically). I think it's possible with Bourbaki's logic. – paeolo Jul 04 '17 at 23:48
  • So make the Banach space be part of the data. Then you cannot have your second atlas. EDIT: Actually, I don't think I agree with you that $X_E$ and $X_F$ should be literally the same object. They're different atlases. They should present different objects in that sense. The objects they present are isomorphic as Banach manifolds. –  Jul 04 '17 at 23:55
  • I don't know much about foundations so forgive me if I'm overreaching, but here's how I'd try to make wikipedia's description legit in a ZFC-like setting: proper classes are really predicates, and likewise an equivalence relation is a predicate with two arguments. The $C^r$-structure on $X$ containing the atlas $A$ can then be seen as the predicate $P(B)$ that tells you whether $B$ is equivalent to $A$. – Anthony Carapetis Jul 04 '17 at 23:57
  • 1
    Really, though, I think this is all meant to just be an intuitive way of thinking about a single structure. In practice, there's no need to construct a $C^r$-structure as a formal mathematical object - a Banach manifold can be defined in terms of a single atlas, and you can introduce refined/compatible atlases as necessary. – Anthony Carapetis Jul 04 '17 at 23:58
  • @MikeMiller Yes but I don't want to avoid the secound atlas, the aim of my question is not to use such restrictions, as it's done in Wikipedia and the book of S. Lang I mentionned. I would prefer a logical solution. – paeolo Jul 05 '17 at 00:05
  • @AnthonyCarapetis I agree that it's not necessary. But I have read many books where it's done without talking about the logic. Except in Bourbaki where it seems to use the Bourbaki's logic, in particular the $\tau$ symbol to pick out a unique representative but I'm not sure how it works.That's why I question myself.. – paeolo Jul 05 '17 at 00:23
  • The issue already occurs in the finite-dimensional case -- you could, if you wanted, allow charts with values in any finite-dimensional vector space. But why would you want to? Everything works out just fine if you insist that all charts take their values in $\mathbb R^n$. The analogous solution for arbitrary Banach manifolds is just to pick one model Banach space and work with that. – Jack Lee Jul 05 '17 at 00:29
  • @JackLee: If we're taking paeolo's approach of allowing mixed dimension, though, the situation seems different to me: $\mathbf {FinVec}$ has a small category of isomorphism class representatives ${ \mathbb R^n : n \in \mathbb N }$ (googling tells me "essentially small" is the terminology I'm looking for), and so if we use these as models then everything's a set. If you try to do this in the general case, the dimensions of Banach spaces form a proper class of cardinals (I think?), so there's no equivalent representative set ("small skeleton"?) of model spaces – Anthony Carapetis Jul 05 '17 at 00:56
  • @AnthonyCarapetis No I think it's still working fine because you start with an atlas which is a set. – paeolo Jul 05 '17 at 00:59
  • @AnthonyCarapetis: OK, I guess. I've never found a need to consider a space whose components have different dimensions as a single manifold. But I'm biased toward the concrete. – Jack Lee Jul 05 '17 at 01:01
  • @JackLee I was already convinced this logical issue is not a problem at all to do differential geometry, as you pointed out. However I have seen this approach many times, with a confident writing, so I was wondering wether it was a known issue, or wether I missed something. – paeolo Jul 05 '17 at 01:07
  • 3
    There's a standard way to handle this in ZF called "Scott's trick". See https://math.stackexchange.com/questions/251958/equivalence-relations-on-classes-instead-of-sets – Eric Wofsey Jul 05 '17 at 01:52
  • @EricWofsey You should turn that into an answer. – Noah Schweber Jul 05 '17 at 15:48

0 Answers0